If $$\sin A = \frac{1}{x}$$ then $$\cos A = ?$$

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RRB NTPC Exam 27 Jan 2021 click here to view all questions

If $$\sin A = \frac{1}{x}$$ then $$\cos A = ?$$

A$$ \frac{\sqrt{x^2+1}}{x} $$

B$$ \frac{x}{\sqrt{x^2-1}} $$

C$$ \frac{\sqrt{x^2-1}}{x} $$

D$$ \frac{x}{\sqrt{x^2+1}} $$

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correct answer is: option c [$\frac{\sqrt{x^2-1}}{x}$] Explanation: We know that $\sin^2 A + \cos^2 A = 1$. Therefore, $\cos^2 A = 1 - \sin^2 A$. Given that $\sin A = \frac{1}{x}$, we can substitute this into the equation to get $\cos^2 A = 1 - (\frac{1}{x})^2 = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}$. Taking the square root of both sides, we get $\cos A = \sqrt{\frac{x^2 - 1}{x^2}} = \frac{\sqrt{x^2 - 1}}{x}$. Option a: $\frac{\sqrt{x^2+1}}{x}$ is incorrect because it uses addition instead of subtraction under the square root. It incorrectly calculates $\cos A$ using a flawed manipulation of the Pythagorean identity. Option b: $\frac{x}{\sqrt{x^2-1}}$ is incorrect because it is the reciprocal of the correct answer, and incorrectly places x in the numerator and the root in the denominator. Option c: $\frac{\sqrt{x^2-1}}{x}$ is the correct answer, as shown in the explanation. Option d: $\frac{x}{\sqrt{x^2+1}}$ is incorrect for two reasons: it contains a '+' instead of a '-', and the numerator and denominator are swapped from the correct response following algebraic manipulation. Akhilesh ? answered Apr 1 '25 at 20:40

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